We can use differentiation to find the derivative, which we have seen has many useful applications. What if we were given a derivative function, can we go backwards to find the original function it came from? Can we "anti-differentiate"?
For example, if we knew that the derivative of some function $f\left(x\right)$f(x) was $6x+4$6x+4, could we find what $f\left(x\right)$f(x) was? What the degree of $f\left(x\right)$f(x) was? Do we have enough information?
Let's explore this problem by first finding the derivative of the following five functions:
| 1. | $F\left(x\right)=3x^2+4x+5$F(x)=3x2+4x+5 |
| 2. | $F\left(x\right)=3x^2+4x+1$F(x)=3x2+4x+1 |
| 3. | $F\left(x\right)=3x^2+4x-17$F(x)=3x2+4x−17 |
| 4. | $F\left(x\right)=3x^2+4x-3$F(x)=3x2+4x−3 |
| 5. | $F\left(x\right)=3x^2+4x+a$F(x)=3x2+4x+a |
What do you notice?
Well, since the constant terms $5,1,-17,-3$5,1,−17,−3 and $a$a all have a derivative of zero, we end up getting the same gradient function for all of them: $F'\left(x\right)=6x+4$F′(x)=6x+4.
So, if we were given that the gradient function of $f\left(x\right)$f(x) is $6x+4$6x+4 and we were looking for the original function, how would we know which one it was? Without extra information, such as a point the original function passes through, we cannot tell what the constant term may have been.
In fact, when we go backwards using anti-differentiation, we cannot be sure which function was the original so we find what is called a primitive function. We can give the form of all possible primitive functions by adding an unknown constant.
For example, all primitive functions of $6x+4$6x+4 have the form $F\left(x\right)=3x^2+4x+C$F(x)=3x2+4x+C, where $C$C is a constant term that we don't yet know the value of. When including the constant we often refer to the form as the primitive function, and this is also known as the general anti-derivative or the indefinite integral.
The process of "going backwards" from differentiation to find a primitive function is called anti-differentiation or indefinite integration.
For power functions of the form $f\left(x\right)=ax^n$f(x)=axn, our differentiation rule tells us $f'\left(x\right)=nax^{n-1}$f′(x)=naxn−1. That is, "Multiply the expression by the power of $x$x and then subtract $1$1 from the power." For anti-differentiation we want to "undo" that. Reversing the rule we get "add $1$1 to the power of $x$x, then divide by the expression by the new power"
So for $f\left(x\right)=ax^n$f(x)=axn, the indefinite integral is $F\left(x\right)=\frac{ax^{n+1}}{n+1}+C$F(x)=axn+1n+1+C. The value $C$C is known as the constant of integration.
Notation: The expression $\int f\left(x\right)\ dx$∫f(x) dx, reads as "anti-differentiate the function $f\left(x\right)$f(x) with respect to $x$x." For example, $\int3x^2dx=x^3+C$∫3x2dx=x3+C, where $C$C is a constant, reads as "the general anti-derivative of $3x^2$3x2 with respect to $x$x is $x^3+C$x3+C".
Worked examples
Example 1
Find the primitive function $F\left(x\right)$F(x) when $f\left(x\right)=3x^2+4x$f(x)=3x2+4x.
Think: Term by term, we raise the power by one and divide by the new power.
Do:
| $F\left(x\right)$F(x) | $=$= | $\frac{3x^{2+1}}{2+1}+\frac{4x^2}{2}+C$3x2+12+1+4x22+C |
Don't forget the constant of integration |
| $=$= | $\frac{3x^3}{3}+2x^2+C$3x33+2x2+C |
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| $=$= | $x^3+2x^2+C$x3+2x2+C, where $C$C is a constant. |
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Example 2
Given that $\frac{dy}{dx}=\frac{1}{x^3}-\sqrt{x}$dydx=1x3−√x, find the form of all anti-derivatives $y$y.
Think: First rewrite all terms as powers of $x$x and then integrate term by term.
Do: $\frac{dy}{dx}=x^{-3}-x^{\frac{1}{2}}$dydx=x−3−x12
| $y$y | $=$= | $\int x^{-3}-x^{\frac{1}{2}}\ dx$∫x−3−x12 dx |
| $=$= | $\frac{x^{-2}}{-2}-\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+C$x−2−2−x32(32)+C | |
| $=$= | $-\frac{1}{2x^2}-\frac{2x^{\frac{3}{2}}}{3}+C$−12x2−2x323+C, where $C$C is a constant. |
Example 3
Given that $f'\left(x\right)=6x^2-4x+1$f′(x)=6x2−4x+1 and $f\left(x\right)$f(x) passes through the point $\left(2,7\right)$(2,7), find the function $f\left(x\right)$f(x).
Think: Find the indefinite integral and then use the given point to find the constant for the given function.
Do:
| $f\left(x\right)$f(x) | $=$= | $\int6x^2-4x+1\ dx$∫6x2−4x+1 dx |
| $=$= | $\frac{6x^3}{3}-\frac{4x^2}{2}+x+C$6x33−4x22+x+C | |
| $=$= | $2x^3-2x^2+x+C$2x3−2x2+x+C |
Given $f\left(x\right)$f(x) passes through $\left(2,7\right)$(2,7), we have:
| $f\left(2\right)$f(2) | $=$= | $7$7 |
| Thus, $2\left(2\right)^3-2\left(2\right)^2+\left(2\right)+C$2(2)3−2(2)2+(2)+C | $=$= | $7$7 |
| $16-8+2+C$16−8+2+C | $=$= | $7$7 |
| $\therefore C$∴C | $=$= | $-3$−3 |
Hence, $f\left(x\right)=2x^3-2x^2+x-3$f(x)=2x3−2x2+x−3.
Practice questions
Question 1
Find the primitive function of $15x^4+16x^3$15x4+16x3.
Use $C$C as the constant of integration.
Question 2
Consider the gradient function $f'\left(x\right)$f′(x)$=$=$4x+3$4x+3.
The family of the antiderivative, $f\left(x\right)$f(x), will be:
Linear
AQuadratic
BCubic
CExponential
DThe form of the antiderivative will be $f\left(x\right)$f(x)$=$=$ax^2+bx+C$ax2+bx+C. State the value of $a$a and $b$b in simplified form if necessary.
$a$a $=$= $\editable{}$
$b$b $=$= $\editable{}$
Which of the following functions could possibly represent $f\left(x\right)$f(x)? Select all that apply.
$f\left(x\right)=4\left(x+3\right)^2+7$f(x)=4(x+3)2+7
A$f\left(x\right)=2x^2+3x$f(x)=2x2+3x
B$f\left(x\right)=2x^2+3x+4$f(x)=2x2+3x+4
C$f\left(x\right)=2x^2+3$f(x)=2x2+3
D
Question 3
Find $\int\left(\frac{x^3}{3}-\frac{3}{x^3}\right)dx$∫(x33−3x3)dx, with $C$C as the constant of integration.
Question 4
Suppose $\frac{dy}{dx}=9x^2-10x+2$dydx=9x2−10x+2.
Find an equation for $y$y.
Use $C$C as the constant of integration.
Solve for $C$C if the curve passes through the point $\left(2,13\right)$(2,13).
Hence find the equation of $y$y.